Multiple Choice 1 point Considering only the values of \( \alpha \) for which \( \tan \alpha \cos (2 \alpha) \) is defined, which of the following expressions is equivalent to tan \( \alpha \cos (2 \alpha) \) ? \( 2 \cos ^{2} \alpha \) \( \cos \alpha \cot \alpha-\sin \alpha \) \( \frac{2 \sec \alpha}{1-\tan ^{2} \alpha} \) \( 2 \sin \alpha \cos \alpha-\tan \alpha \) \( 2 \sin ^{2} \alpha-\tan ^{2} \alpha \)
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Mind Expander
To determine which expression is equivalent to \( \tan \alpha \cos (2 \alpha) \), we can first rewrite \( \tan \alpha \) and \( \cos(2\alpha) \): \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha}, \quad \cos(2\alpha) = \cos^2 \alpha - \sin^2 \alpha = 2\cos^2 \alpha - 1 \] Thus, \[ \tan \alpha \cos(2 \alpha) = \frac{\sin \alpha}{\cos \alpha} (2 \cos^2 \alpha - 1) = \sin \alpha \left( \frac{2 \cos^2 \alpha - 1}{\cos \alpha} \right) = \sin \alpha (2 \cos \alpha - \tan \alpha) \] Now let's look at the options: 1. \( 2 \cos^2 \alpha \) - does not represent \( \tan \alpha \cos(2\alpha) \). 2. \( \cos \alpha \cot \alpha - \sin \alpha = \frac{\cos \alpha}{\tan \alpha} - \sin \alpha = 0 \) is not equal to \( \tan \alpha \cos(2\alpha) \). 3. \( \frac{2 \sec \alpha}{1-\tan^2 \alpha} = \frac{2/\cos \alpha}{1 - \sin^2/\cos^2} = 2\cos^2\alpha \), not equivalent. 4. \( 2 \sin \alpha \cos \alpha - \tan \alpha = \sin(2\alpha) - \tan \alpha \) 5. \( 2 \sin^2 \alpha - \tan^2 \alpha = 2 \sin^2 \alpha - \frac{\sin^2 \alpha}{\cos^2 \alpha} \) The correct answer is \( \cos \alpha \cot \alpha - \sin \alpha \).
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